Question

Use z scores to compare the given values.
Based on sample​ data, newborn males have weights with a mean of 3217.9g and a standard deviation of 780.1g. Newborn females have weights with a mean of 3086.5g and a standard deviation of 523.8g. Who has the weight that is more extreme relative to the group from which they​ came: a male who weighs 1700g or a female who weighs ​1700g?

Answer 1

Using z-scores, it is found that a female who weighs ​1700g has the more extreme value relative to the group from which they came.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

For the newborn male, we have that $X = 1700, \mu = 3217.9, \sigma = 780.1$, hence the z-score is given by:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1700 - 3217.9}{780.1}$

Z = -1.946.

For the newborn female, we have that $X = 1700, \mu = 3086.5, \sigma = 523.8$, hence the z-score is given by:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1700 - 3086.5}{523.8}$

Z = -2.645.

Due to the higher absolute value of the z-score, a female who weighs ​1700g has the more extreme value relative to the group from which they ca,e.

More can be learned about the z-scores distribution at https://brainly.com/question/24663213

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